Students can access the CBSE Sample Papers for Class 10 Maths Basic with Solutions and marking scheme Set 8 will help students in understanding the difficulty level of the exam.
CBSE Sample Papers for Class 10 Maths Basic Set 8 with Solutions
Time: 3 Hours
Maximum Marks: 80
General Instructions:
1. This Question Paper has 5 Sections A, B, C, D and E.
2. Section A has 20 Multiple Choice Questions (MCQs) carrying 1 mark each.
3. Section B has 5 Short Answers-I (SA-I) type questions carrying 2 marks each.
4. Section C has 6 Short Answers-II (SA-II) type questions carrying 3 marks each.
5. Section D has 4 Long Answers (LA) type questions carrying 5 marks each.
6. Section E has 3 source-based/case-based/passage-based integrated units of assessment (4 marks each) with sub-parts of the values of 1, 1, and 2 marks each respectively.
7. All questions are compulsory. However, an internal choice in 2 Qs of 2 marks, 2 Qs of 3 marks, and 2 Qs of 5 marks has been given provided. An internal choice has been provided in 2-mark questions of Section E.
8. Draw neat figures wherever required, Take. x = 2, wherever required if not stated.
Section-A(20 Marks)
Question 1.
1245 is a factor of the numbers p and q.
Which of the following will always have 1245 as a factor?
(i) p + q
(ii) p × q
(iii) p / q
(A) only (ii)
(B) only (i) and (ii)
(C) only (ii) and (iii)
(D) all – (i), (ii) and (iii)
Solution:
(B) only (i) and (ii)
Explanation: Only p + q and p × q have 1245 as a factor.
Question 2.
The (HCF × LCM) for the numbers 50 and 20 is
(A) 1000
(B) 50
(C) 100
(D) 500
Solution:
(A) 1000
Explanation: We know that HCF × LCM
= Product of two numbers
⇒ HCF × LCM = 20 × 50
∴ HCF × LCM = 1000
Question 3.
If 1/2 is a root of the equation x2 + kx – 5/4 = 0, then the value of k is
(A) 2
(B) -2
(C) 1/2
(D) 1/2
Solution:
(A) 2
Explanation: Since, 1/2 is a root of the equation
Question 4.
The value of k for which the equations 3x – y + 8 = 0 and 6x + ky = -16 represent coincident lines, is
(A) -3
(B) 2
(C) 2
(D) -2
Solution:
(D) -2
Explanation: For Coincident lines the pair of linear equations has infinitely many solutions
\(\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\)
Here, a1 = 3, a2 = 6, b1 = –1, b2 = k, c1 = 8, c2 =16
So, we have \(\frac{3}{6}=\frac{-1}{k}\) ⇒k = -2
Question 5.
The number of quadratic polynomials having 2 and -5 as their zeroes are
(A) 1
(B) 2
(C) 3
(D) infinite
Solution:
(D) infinite
Explanation: Since, we can obtain many polynomials by multiplying or dividing the coefficient of polynomial with numeric constants.
So, there are infinite quadratic polynomials with given two zeroes.
Question 6.
If point P(6, 2) divides the line segment joining A(6, 5) and B(4, y) in the ratio 3: 1, then the value of y is
(A) 4
(B) 3
(C) 2
(D) 1
Solution:
(D) 1
Explanation:
Here, x1 = 6, y1 = 5
Question 7.
In the following figure, Q is a point on PR and S is a point on TR. QS is drawn and ∠RPT = ∠RQS.
Which of these criteria can be used to prove that ΔRSQ is similar to ΔRTP?
(A) AAA similarity criterion
(B) SAS similarity criterion
(C) SSS similarity criterion
(D) RHS similarity criterion
Solution:
(A) AAA similarity criterion
In DRSQ and ∠RTP,
∠R = ∠R (common)
∠RPT = ∠RQS (Given)
By AA similarity rule,
DRSQ ~ DRTP
When two angles are equal of two triangle then third angle is also equal, therefore we can say both triangles are similar by AAA similarity.
Question 8.
In the given figure, if DE || BC. Find EC.
(A) 1 cm
(B) 2 cm
(C) 3 cm
(D) 4 cm
Solution:
(B) 2 cm
\(\begin{aligned}
\frac{A D}{B D} & =\frac{A E}{E C} \\
\frac{1.5}{3} & =\frac{1}{E C}
\end{aligned}\)
∴ EC = 2 cm
Question 9.
If the radii of two concentric circles are 4 cm and 5 cm, then the length of each chord of one circle which is the tangent to the other circle is
(A) 3 cm
(B) 6 cm
(C) 9 cm
(D) 1 cm
Solution:
(B) 6 cm
Explanation: Let C1, C2 be two concentric circles with their centre C.
Let, chord AB of circle C2 touches C1 at P
AB is tangent at P and PC is radius
∴ CP ⊥ AB
Given, ∠P = 90°, CP = 4 cm and CA = 5 cm
In right angle ΔPAC,
AP2 = AC2 – PC2
= 52 – 42
= 25 – 16
= 9
∴ AP = 3 cm
Q Perpendicular from centre to chord bisects the chord.
∴ AB = 2AP
= 2 × 3 = 6
Question 10.
if tan (3x + 30°) = 1, then find the value of x.
(A) 2
(B) 3
(C) 4
(D) 5
Solution:
(D) 5
Explanation: tan(3x + 30°) = 1
Or, tan(3x + 30°) = tan 45°
Or, 3x + 30° = 45°
Or, 3x = 45° – 30°
Or, 3x = 15°
Or, x = 15°/3 = 5°
Question 11.
If the length of the ladder placed against a wall is twice the distance between the foot of the ladder and the wall. Find the angle made by the ladder with the horizontal.
(A) 60°
(B) 45°
(C) 30°
(D) None of these
Solution:
(A) 60°
Let the distance between the foot of the ladder and the wall, AB be x.
Then AC, the length of the ladder = 2x
Question 12.
Evaluate (1 – tan 45° + cosec 30°)/(1 + tan 45°).
(A) 1
(B) 2
(C) 3
(D) 0
Solution:
(A) 1
(1 – tan 45 + cosec 30)/(1 + tan 45 )
= (1 – 1 + 2)/(1 + 1)
= 2/2
= 1
Question 13.
It proposed to build a single circular park equal in area to the sum of areas of two circular parks of diameters 16 m and 12 m in a locality. The radius of the new park would be
(A) 10m
(B) 15m
(C) 20m
(D) 24m
Solution:
(A) 10m
Explanation: Area of first circular park whose diameter is 16 m,
\(\begin{aligned}
& =\pi\left(\frac{16}{2}\right)^2 \\
& =\pi(8)^2
\end{aligned}\)
= 64 πm2
Area of second circular park whose diameter is 12 m,
\(\begin{aligned}
& =\pi\left(\frac{12}{2}\right)^2 \\
& =\pi(6)^2
\end{aligned}\)
= 36 πm2
According to question,
Area of single circular park
= Area of first circular park + Area of second circular park
πr2 = 64π + 36π
πr2 = 100π
r = 10 m
Question 14.
if the perimeter and the area of a circle are numerically equal, then the radius of the circle is
(A) 2 units
(B) π units
(C) 4 units
(D) 2π units
Solution:
(A) 2 units
Explanation: Given that,
⇒ 2πr = πr2
⇒ r = 2 units
Question 15.
A die is thrown once. Find the probability of getting “at most 2”.
(A) \(\frac{1}{6}\)
(B) \(\frac{2}{3}\)
(C) \(\frac{1}{2}\)
(D) \(\frac{1}{3}\)
Solution:
(D) \(\frac{1}{3}\)
S = {1, 2, 3, 4, 5, 6}
n(S) = 6
A = {1, 2}
n(A) = 2
∴ P(A) = \(\frac{n(A)}{n(S)}=\frac{2}{6}=\frac{1}{3}\)
Question 16.
The weights of tea in 70 packets are shown in the following table:
Weight(in grams) | Number of packets |
200-201 | 13 |
201-202 | 27 |
202-203 | 18 |
203-204 | 10 |
204-205 | 1 |
205-206 | 1 |
The modal class is
(A) 200-201
(B) 201-202
(C) 205-206
(D) None of these
Solution:
(B) 201-202
Explanation: The class interval with highest frequency is 201 – 202
Therefore, the modal class is 201-202.
Question 17.
is the length of arc of a circle of radius 7 cm which subtends an angle of 90° at the centre of the circle?
(A) 22 cm
(B) 11 cm
(C) \(\frac{77}{2}\) cm
(D) \(\frac{11}{2}\) cm
Solution:
(B) 11cm
Explanation: Given that,
Radius of circle = 7 cm
Central angle = 90°
Length of arc = 2πr\(\left(\frac{90^{\circ}}{360^{\circ}}\right)\)
\(\begin{aligned}
& =\pi \frac{r}{2} \\
& =\frac{22}{7} \times \frac{7}{2}
\end{aligned}\)
= 11 cm
Question 18.
The mean of first ten natural numbers is
(A) 5.5
(B) 55
(C) 45
(D) 45
Solution:
(A) 5.5
Explanation: \(\frac{1+2+3+4+5+6+7+8+9+10}{10}=\frac{55}{10}\) = 5.5
Directions: In the following questions, A Statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as.
Question 19.
Assertion (A): A number N when divided by 15 gives the remainder 2. Then the remainder is same when N is divided by 5.
Reason (R): √3 is an irrational number.
(A) Both (A) and (R) are true and (R) is the correct explanation for (A).
(B) Both (A) and (R) are true but (R) is not the correct explanation for (A).
(C) (A)is true but (R) is false.
(D) (A) is false but (R) is true.
Solution:
(B) Both (A) and (R) are true but (R) is not the correct explanation for (A).
Explanation: If we divide 32, 47, 62 by 15 and 5 both remainder is always 2.
∴ Both Assertion and Reason are true.
But Reason is not the correct explanation of assertion.
Question 20.
Assertion (A): The point (-1, 6) divides the line segment joining the points (-3, 10) and (6, -8) in the ratio 2:7 internally.
Reason (R): Three points are A, B and C are collinear if area ΔABC = 0.
(A) Both (A) and (R) are true and (R) is the correct explanation for (A).
(B) Both (A) and (R) are true but (R) is not the correct explanation for (A).
(C) (A)is true but (R) is false.
(D) (A)is false but (R) is true.
Solution:
(B) Both (A) and (R) are true but (R) is not the correct explanation for (A).
Explanation: In case of assertion:
Let the ratio at which P divides AB is k: 1.
On solving we get, k = 2/7 Which is true.
In case of Reason:
Collinear points are the points which lie on the straight line so making a triangle is not possible.
So, Area of triangle = 0.
Therefore, Both A and R are true and R is not correct explanation for A.
Section -B(10 Marks)
Question 21.
Find 21st term of the A.P, whose first two terms are -3 and 4.
Solution:
Given,
t1 = -3 and t2 = 4
⇒ d = t2 – t1 = 4 – (-3) = 7
tn = a + (n – 1 )d
⇒ t21 = (-3) + (21 – 1)(7) = 137
Question 22.
In the adjoining figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR.
Show that BC || QR.
Solution:
OR
A vertical pole of length 15 m casts a shadow 10 m long on the ground and at the same time a tower casts a shadow 48 m long. Find the length of the tower.
Solution:
Let height of the pole be AB and length of is shadow is BC
Similarly, Height of the tower is DE and length of its shadow is EF.
In ΔABC & ΔDEF,
∠B = ∠E = 90°
∠C = ∠F [Angle of elevation in both cases are equal at the same time]
BY AA Similarity rule,
ΔABC ~ ΔDEF
Therefore, \(\frac{A B}{D E}=\frac{B C}{E F}\)
⇒ \(\begin{aligned}
\frac{15}{h} & =\frac{10}{48} \\
h & =\frac{15 \times 48}{10}
\end{aligned}\) = 72 m
Question 23.
In the given figure, from a point P, two tangents PT and PS are drawn to a circle with centre O such that ∠SPT = 120°, prove that OP = 2PS.
Solution:
Question 24.
Prove that: \(\frac{\left(\sin ^4 \theta+\cos ^4 \theta\right)}{1-2 \sin ^2 \theta \cos ^2 \theta}\) = 1
Solution:
Question 25.
The minute hand of a clock is 12 cm long. Find the area of the face of the clock described by the minute hand in 35 minutes.
Solution:
Length of minute hand = radius of the sector formed
by minute hand = 12 cm
Total time = 35 minutes.
Angle covered by minute hand in 1 minute
= \(\frac{360}{60}\) = 6°
Angle covered by minute hand in 35 minutes
= 35 × 6
= 210°
Area of the face of the clock described by minute hand in 35 minutes
= Area of sector
\(\begin{aligned}
& =\frac{\theta}{360} \pi r^2 \\
& =\frac{210}{360} \times \frac{22}{7} \times 12 \times 12
\end{aligned}\)
= 264 cm²
OR
A wire is looped in the form of a circle of radius 35 cm. It is rebent into a square form. Find the length of side of the square.
Solution:
Radius of the circle = 35 cm
Length of wire = circumference of the circle
= 2 × \(\frac{22}{7}\) × 35
= 44 × 5 = 220 cm
Therefore, perimeter of the square = length of wire = 220 cm
Hence, the side of the square
\(\begin{aligned}
& =\frac{(\text { Perimeter of the square })}{4} \\
& =\frac{220}{4}=55 \mathrm{~cm}
\end{aligned}\)
Section -C(18 Marks)
Question 26.
Enlist the types of real numbers and define rational number. Also prove that √3 is an irrational number.
Solution:
Various types of real numbers are:
- Rational number
- Irrational number
- Whole number
- Integer
- Natural number
Rational number: Any number that can be expressed in the form of \(\frac{p}{q}\) where q ≠ 0 and p, q are integers is called a rational number.
Let √3 be a rational number
∴ √3 = \(\frac{p}{q}\), where p and q are co–prime integers and q ≠ 0.
On squaring both sides, we get 3 = \(\frac{p^2}{q^2}\)
Or, p2 = 3q2
∴ p2 is divisible by 3.
∴ p is divisible by 3. …(i)
Let p = 3r for some positive integer r.
∴ p2 = 9r2
Or, 3q2 = 9r2
Or, q2 = 3r2
Or, q2 is divisible by 3.
∴ q is divisible by 3. …(ii)
From eqn. (i) and (ii), p and q are divisible by 3, which contradicts the fact that p and q are coprimes.
Hence, our assumption is wrong.
∴ √3 is an irrational number.
Question 27.
If α and β are the zeroes of the polynomial f (x) = x2 – 4x – 5 then find the value of α2 + β2.
Solution:
Question 28.
A fraction becomes \(\frac{1}{3}\) when 2 is subtracted from the numerator and it becomes \(\frac{1}{2}\) when 1 is subtracted from the denominator. Find the fraction.
Solution:
Let the fraction be x/y
∴ \(\frac{x-2}{y}=\frac{1}{3}\frac{x}{y-1}=\frac{1}{2}\)
Solving (i) and (ii) to get x = 7, y = 15.
∴ Required fraction is 7/15
Detailed Solution :
Let the fraction be x/y.
According to the first condition, \(\frac{x-2}{y}=\frac{1}{3}\)
⇒ 3x – 6 = y
⇒ y = 3x – 6 …(i)
According to the second condition, \(\frac{x}{y-1}=\frac{1}{2}\)
⇒ 2x = y – 1
⇒ y = 2x + 1 …(ii)
From eq’s. (i) and (ii), we get
3x – 6 = 2x + 1
⇒ x = 7
Substitute value of x in eq (i), we get
y = 3(7) – 6
y = 21 – 6 = 15
Hence, the fraction is 7/15
OR
5 pencils and 7 pens together cost ₹ 250 whereas 7 pencils and 5 pens together cost ₹ 302. Find the cost of one pencil and that of a pen.
Solution:
Let the cost of a pencil = x and cost of a pen = y.
As per given conditions,
5x + 7y = 250 (1)
7x + 5y = 302 (2)
Multiplying (1) with 7 and (2) with 5
35x + 49y = 1750 ……(3)
35x + 25y = 1510
(3)-(4) gives
24y = 240
y = 10
Substitute y = 10 in (1)
5x + (7 × 10) = 250
5x = 250 – 70
x = 36
Cost of a pencil = Rs. 36
Cost of a pen = Rs. 10
Question 29.
In the given figure, PA and PB are tangents to a circle from an external point P such that PA = 4 cm and ∠BAC = 135°, Find the length of chord AB.
Solution:
PA = PB = 4 cm [Tangents from external point]
∠PAB = 180° – 135° = 45° [Supplementary angles]
∠ABP = ∠PAB = 45° [Opposite angles of equal sides]
∴ ∠APB = 180° – 45° – 45°
= 90°
So, ∠ABP is an isosceles right angled triangle.
⇒ AB2 = 2AP2
⇒ AB2 = 32
Hence, AB = √32 = 4√2 cm
Question 30.
Prove that: \(\frac{\cot \theta+\operatorname{cosec} \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}=\frac{1+\cos \theta}{\sin \theta}\)
Solution:
OR
Prove that: sin θ (1 + tan θ) + cos θ (1 + cot θ) = sec θ + cosecθ.
Solution:
Question 31.
In a single throw of a pair of different dice, what is the probability of getting (i) a prime number on each dice? (ii) a total of 9 or 11?
Solution:
(i) Favourable outcomes are (2, 2) (2, 3) (2, 5) (3, 2) (3, 3) (3, 5) (5, 2) (5, 3) (5, 5) i.e., 9 outcomes.
P(a prime number on each die) = \(\frac{9}{36} \text { or } \frac{1}{4}\)
(ii) Favourable outcomes are (3, 6) (4, 5) (5, 4) (6, 3) (5, 6) (6, 5) i.e., 6 outcomes
P(a total of 9 or 11) = \(\frac{6}{36} \text { or } \frac{1}{6}\)
Section-D(20 Marks)
Question 32.
The ninth term of an AP is equal to seven times the second term and the twelfth term exceeds five times the third term by 2. Find the first term and the common difference.
Solution:
Let the first term of A.P. be a and common difference be d.
Given, a9 = 7a2
Or, a + 8d = 7(a + d) …(i)
and a12 = 5a3 + 2
Again, a + 11d = 5(a + 2d) + 2 …(ii)
From (i), a + 8d = 7a + 7d
– 6a + d = 0 …(iii)
From (ii), a + 11d = 5a + 10d + 2
– 4a + d = 2 …(iv)
Subtracting (iv) from (iii), we get
– 2a = – 2
Or, a = 1
From (iii),
– 6 + d = 0
d = 6
Hence, first term = 1 and common difference = 6
OR
Find the value of a, b and c such that the numbers a, 7, b, 23 and c are in A.P
Solution:
Since, a, 7, b, 23 and c are in A.P.
Let the common difference be d
∴ a + d = 7 …(i)
and a + 3d = 23 …(ii)
From (i) and (ii), we get
a = – 1 and d = 8
Again, b = a + 2d
b = – 1 + 2 × 8
Or, b = – 1 + 16
Or, b = 15
∴ c = a + 4d
= – 1 + 4 × 8
= – 1 + 32
c = 31
∴ a = – 1, b = 15 and c = 31
Question 33.
AB is a chord of circle with centre O. At B, a tangent PB is drawn such that its length is 24 cm. The distance of P from the centre is 26 cm. If the chord AB is 16 cm, find its distance from the centre.
Solution:
Given, AB is a chord of circle with centre O and tangent PB = 24 cm, OP = 26 cm.
Construction: Join O to B and draw OC ⊥ AB.
By Pythagoras theorem
\(\begin{aligned}
O B & =\sqrt{(26)^2-(24)^2} \\
& =\sqrt{676-576}=\sqrt{100}
\end{aligned}\) = 10 cm
Now, in ΔOBC, BC = 1/2 AB = 16/2 = 8 cm
(Perpendicular drawn from the centre to a chord bisects it.)
OB = 10 cm
OC2 = OB2 – BC2
= 102 – 82
OC2 = 36
OC = 6 cm
∴ Distance of the chord from the centre = 6 cm
Question 34.
A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.
Solution:
OR
Ramesh made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath.
Solution:
Let h be height of the cylinder, and r the common
radius of the cylinder and hemisphere.
Then, the total surface area = CSA of cylinder + CSA of hemisphere
= 2prh + 2pr2 = 2pr (h + r)
Question 35.
The distribution below gives the weights of 30 students of a class. Find the median weight of the students.
Solution:
Weight in kg | Number of Students (f) | Cumulative Frequency (Cf) |
40-45 | 2 | 2 |
45-50 | 3 | 5 |
50-55 | 8 | 13 |
55-60 | 6 | 19 |
60-65 | 6 | 25 |
65-70 | 3 | 28 |
70-75 | 2 | 30 |
Section -E(12 Marks)
Question 36.
Lokesh, a production manager in Mumbai, hires a taxi everyday to go to his office. The taxi charges in Mumbai consist of a fixed charge together with the charges for the distance covered. His office is at a distance of 10 km from his home. For a distance of 10 km to his office, Lokesh paid ₹ 105. While coming back home, he took another route. He covered a distance of 15 km and the charges paid by him were ₹ 155.
Based on the above information, answer the following questions:
(i) What are the fixed charges?
(ii) What are the charges per km?
(iii) If fixed charges are ₹ 20 and charges per km are ₹ 10, then how much Lokesh have to pay for traveling a distance of 10 km?
Solution:
Let the fixed charge be ₹ x and per kilometer charge be ₹ y
∴ x + 10y = 105 …(i)
x + 15y = 155 …(ii)
from (i) & (ii)
5y = 50
∴ y = 50/5 = 10
from eqn. (i) x + 100 = 105
x = 105 – 100 = 5
(i) Fixed charges = ₹ 5
(ii) Per km charges = ₹ 10
(iii) a + 10b
20 + 10 × 10 = ₹ 120
OR
Find the total amount paid by Lokesh for travelling 10 km from home to office and 25 km from office to home. [Fixed charges and charges per km are as in (i) & (ii).
Solution:
Total amount = x + 10y + x + 25y
= 2x + 35y
= 2 × 5 + 35 × 10
= 10 + 350
= ₹ 360
Question 37.
In a GPS, the lines that run east-west are known as lines of latitude, and the lines running north-south are known as lines of longitude. The latitude and the longitude of a place are its coordinates and the distance formula is used to find the distance between two places. The distance between two parallel lines is approximately 150 km. A family from Uttar Pradesh planned a round trip from Lucknow (L) to Puri (P) via Bhuj (B) and Nashik (N) as shown in the given figure below.
Based on the above information answer the following questions using the coordinate geometry.
(i) Find the distance between Lucknow (L) to Bhuj(B).
(ii) If Kota (K), internally divide the line segment joining Lucknow (L) to Bhuj (B) into 3: 2, then find the coordinate of Kota (K).
(iii) Name the type of triangle formed by the places Lucknow (L), Nashik (N) and Puri (P)
Solution:
OR
Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri(P).
Solution:
Let A (0, b) be a point on the y-axis, then AL = AP.
⇒ \(\sqrt{(5-0)^2+(10-b)^2}=\sqrt{(8-0)^2+(6-b)^2}\)
⇒ (5)2 + (10 – b)2 = (8)2 + (6 – b)2
⇒ 25 + 100 – 20b + b2 = 64 + 36 – 12b + b2
⇒ 8b = 25 ⇒ b = 25/8
So, the coordinate on y – axis is (0, 25/3)
Question 38.
‘Skysails’ is that genre of engineering science that uses extensive utilization of wind energy to move a vessel in the sea water. The sky sails technology allows the towing kite to gain a height of anything between 100 m to 300 m. The sailing kite is made in such a way that it can be raised to its proper elevation and then brought back with the help of a telescopic mast that enables the kite to be raised properly and effectively.
(i) Find the value of tan 30°. cot 60°.
(ii) What should be the length of the rope of the kite sail in order to pull the ship at the angle θ and be at a vertical height of 200 m?
(iii) If cos A = 1/2 then find the value of 9 cot2 A – 1
Solution:
OR
Find a place (point) on the longitude (y-axis) which is equidistant from the points Lucknow (L) and Puri(P).
Solution: